6. 下列计算正确的是 ()

A.$\sqrt{\frac{5}{3}}=\frac{1}{3}\sqrt{5}$
B.$\sqrt{\frac{8}{4}} = 2$
C.$\sqrt{\frac{a}{4b}}=\frac{1}{2b}\sqrt{a}$
D.$\sqrt{\frac{5}{4}}=\frac{1}{2}\sqrt{5}$

6. D

7. 化简$\sqrt{2}÷(\sqrt{2}-1)$的结果是 ()

A.$2\sqrt{2}-1$
B.$2-\sqrt{2}$
C.$1-\sqrt{2}$
D.$2+\sqrt{2}$

7. D

8. 计算：

(1) $\frac{\sqrt{5}×\sqrt{15}}{\sqrt{3}}$；
(2) $\sqrt{18}÷\sqrt{\frac{3}{4}}×\sqrt{\frac{4}{3}}$；
(3) $\sqrt{27}×\frac{\sqrt{2}}{3}÷\sqrt{6}$；
(4) $\sqrt{15}×\frac{3}{5}\sqrt{20}÷(-\sqrt{6})$.

8. (1)解：原式$=\dfrac{\sqrt{5}×\sqrt{5}×\sqrt{3}}{\sqrt{3}}=5$
(2)解：原式$=\sqrt{18×\dfrac{4}{3}×\dfrac{4}{3}}=4\sqrt{2}$
(3)解：原式$=3\sqrt{3}×\dfrac{\sqrt{2}}{3}÷\sqrt{6}$
$=\sqrt{6}÷\sqrt{6}=1$
(4)解：原式$=\dfrac{3}{5}×\sqrt{15×20}÷(-\sqrt{6})$
$=\dfrac{3}{5}×\sqrt{300}÷(-\sqrt{6})$
$=\dfrac{3}{5}×10\sqrt{3}÷(-\sqrt{6})$
$=6\sqrt{3}÷(-\sqrt{6})=-6×\sqrt{\dfrac{3}{6}}$
$=-6×\sqrt{\dfrac{1}{2}}=-6×\dfrac{\sqrt{2}}{2}=-3\sqrt{2}$

1. 设长方形的面积为$S$，相邻两边长分别为$a$，$b$，已知$S = 6\sqrt{5}$，$b = \sqrt{15}$，求$a$的值.

1. 解：$\because S=ab$，
$\therefore a=\dfrac{S}{b}=\dfrac{6\sqrt{5}}{\sqrt{15}}=6×\sqrt{\dfrac{5}{15}}=6×\sqrt{\dfrac{1}{3}}$
$=6×\dfrac{\sqrt{3}}{3}=2\sqrt{3}$.
答：$a=2\sqrt{3}$.

2. 已知长方体体积$V = 8\sqrt{5}$，高$h = 2\sqrt{3}$，求它的底面积$S$.

2. 解：$\because V=Sh$，$\therefore S=\dfrac{V}{h}$
$=\dfrac{8\sqrt{5}}{2\sqrt{3}}=4×\sqrt{\dfrac{5}{3}}=4×\dfrac{\sqrt{15}}{3}=\dfrac{4\sqrt{15}}{3}$.
答：它的底面积为$\dfrac{4\sqrt{15}}{3}$.

3. 观察下列各式：①$\sqrt{1+\frac{1}{3}} = 2\sqrt{\frac{1}{3}}$；②$\sqrt{2+\frac{1}{4}} = 3\sqrt{\frac{1}{4}}$；③$\sqrt{3+\frac{1}{5}} = 4\sqrt{\frac{1}{5}}$；…；第9个式子为. 用含$n$（$n$为正整数）的式子写出你猜想的规律：.

3. $\sqrt{9+\dfrac{1}{11}}=10\sqrt{\dfrac{1}{11}}$
$\sqrt{n+\dfrac{1}{n+2}}=(n+1)\sqrt{\dfrac{1}{n+2}}$